PDF of Maximum of Uniform Variables

Solution & Explanation

The task is to derive the Probability Density Function (PDF) for $Z = \max(X, Y)$ given that $X$ and $Y$ are uniformly distributed over the interval $(0,1)$. This involves several steps, including understanding the cumulative distribution function (CDF) and then deriving the PDF from it.

Step 1: Understanding the CDF of Uniform Distribution

Given that both $X$ and $Y$ are uniformly distributed over $(0, 1)$, their CDF is defined as:

• $F_X(x) = P(X \leq x) = x$ for $0 \leq x \leq 1$
• $F_X(x) = 0$ for $x < 0$
• $F_X(x) = 1$ for $x > 1$

The same applies to $Y$ since it is identically distributed.

Step 2: Finding the CDF of $Z = \max(X, Y)$

To find the CDF of $Z$, we need to compute $P(Z \leq z)$. Since $Z$ is the maximum of $X$ and $Y$, the event $Z \leq z$ is equivalent to both $X \leq z$ and $Y \leq z$ occurring simultaneously. Hence:

$P(Z \leq z) = P(X \leq z \cap Y \leq z) = P(X \leq z) \cdot P(Y \leq z)$

Given the independence of $X$ and $Y$, and using their CDF:

$P(Z \leq z) = z \cdot z = z^2 \quad \text{for} \quad 0 \leq z \leq 1$

Step 3: Deriving the PDF from the CDF

The PDF of a random variable is the derivative of its CDF with respect to $z$:

$f_Z(z) = \frac{d}{dz} F_Z(z) = \frac{d}{dz} (z^2) = 2z \quad \text{for} \quad 0 \leq z \leq 1$

For values of $z$ outside $(0, 1)$, the PDF is zero:

• $f_Z(z) = 0$ for $z < 0$ or $z > 1$

Conclusion

The PDF of $Z = \max(X, Y)$, where $X$ and $Y$ are independent and uniformly distributed over $(0,1)$, is given by:

$f_Z(z) = \begin{cases} 2z, & \text{if } 0 \leq z \leq 1 \\ 0, & \text{otherwise} \end{cases}$

This solution demonstrates the critical steps in deriving the PDF from the CDF and highlights the properties of uniform distributions and maxima of random variables.